min/max point
minimum or maximum
f(x)=2x^2+8x-2
a=2
x=(-b)/(2a)=(-8)/(2)(2)=-4
x=-4
y=2(-4)^2+8(-4)-2
2(16)-32-2
32-32-2
y=-2
(-4,-2)
axis of symmetry is x=-4
range[-2,oo]
minimum value -2
3 answers
You have the correct min.
Isn't this the same question as this one?
w-ww-.-jiskha.-c-om/display-.cgi?id=-1207258851
Err exactly the same. The min is wrong >.<. The minimum value occurs at x = -2 but the minimum value is not -2.
w-ww-.-jiskha.-c-om/display-.cgi?id=-1207258851
Err exactly the same. The min is wrong >.<. The minimum value occurs at x = -2 but the minimum value is not -2.
Your parabola in standard form is
y = 2(x+2)^2 - 10 , (check it by expanding this)
so you have a parabola opening upwards with vertex at (-2,-10)
Minimum value is -10, when x=-2
axis of symmetry is x=-2
domain: any real number for x
range: y ≥ -10, y any real number
y = 2(x+2)^2 - 10 , (check it by expanding this)
so you have a parabola opening upwards with vertex at (-2,-10)
Minimum value is -10, when x=-2
axis of symmetry is x=-2
domain: any real number for x
range: y ≥ -10, y any real number